Problem: Find the quadratic polynomial $p(x)$ such that $p(-3) = 10,$ $p(0) = 1,$ and $p(2) = 5.$
Solution: Let $p(x) = ax^2 + bx + c.$  Then from the given information,
\begin{align*}
9a - 3b + c &= 10, \\
c &= 1, \\
4a + 2b + c &= 5.
\end{align*}Then $9a - 3b = 9$ and $4a + 2b = 4,$ which reduce to $3a - b = 3$ and $2a + b = 2.$  Adding, we get $5a = 5,$ so $a = 1.$  Then $4 + 2b = 4,$ so $b = 0.$  Therefore, $p(x) = \boxed{x^2 + 1}.$